Question

Solve It ! : Challenging ​

Answer 1

In ΔACB and ΔCDB

∠ABC = ∠DBC (coincidence)

∠ACB = ∠CDB (90° each)

Hence, ΔACB ~ ΔCDB (AA criterion of similarity)

⇒ ∠DCB = ∠CAB (similar triangles are equiangular)

⇒ AC/CD = CB/DB = AB/CB (ratio of corresponding sides of similar traingles are equal)

Again, in ΔACB and ΔCED,

∠ACB = ∠CED (90° each)

∠CAB = ∠ECD (showed above)

⇒ ΔACB ~ ΔCED (AA)

Also, ΔACB ~ ΔCDB

⇒ ΔCDB ~ ΔCED

SImilarly, we can find all similar triangles and see that

ΔACB ~ ΔADC ~ ΔCDB ~ ΔCED

Now, from ΔADC ~ CED  we get

AC/CD = DC/ED

⇒ CD² = AC × DE

From ΔACB ~ ΔADC

AC/AD = AB/AC

⇒ AC² = AB × AD

Multiplying the two, we get

CD² × AC² = AC × DE × AB × AD

⇒ CD² × AC²/AC = AB × AD × DE

⇒ CD² × AC = AB × AD × DE

Hence proved.